In the previous section, we saw power derived from a torque. In this section, we will see .angular momentum in case of rotation about a fixed axis'
In chap 7.17, we saw the basics about angular momentum of a system of particles
We saw the following information:
Information 1:
Eq.7.19: $\mathbf\small{\frac{dl}{dt}=\vec{r}\times \vec{F}=\vec{\tau}}$
• That is: 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
Information 2:
• To get the total angular momentum $\mathbf\small{\vec{L}}$ of a system of particles, we add the angular momenta of individual particles
• Thus for a system of n particles, we have:
$\mathbf\small{\vec{L}=\vec{l}_1+\vec{l}_1+\vec{l}_1\;+\; .\; .\; .\;+\vec{l}_n=\sum\limits_{i=1}^{i=n}{\vec{l}_i}}$
• Note that, this is an addition of vectors. That is., we have to add the momenta vectorially
• In the above equation, we know that: $\mathbf\small{\vec{l}_i=\vec{r}_i \times \vec{p}_i}$
• So the equation in (1) becomes:
Eq.7.20: $\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
• If required, the $\mathbf\small{\vec{p}_i}$ can be further split up as: $\mathbf\small{\vec{p}_i=m_i \times \vec{v}_i}$
Information 3:
Eq.7.21: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_{ext}}$
• Based on this equation, we can write:
The 'rate of change of angular momentum of a system of particles' at any instant is equal to the 'sum of external torques on all the particles'
In chap 7.17, we saw the basics about angular momentum of a system of particles
We saw the following information:
Information 1:
Eq.7.19: $\mathbf\small{\frac{dl}{dt}=\vec{r}\times \vec{F}=\vec{\tau}}$
• That is: 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
Information 2:
• To get the total angular momentum $\mathbf\small{\vec{L}}$ of a system of particles, we add the angular momenta of individual particles
• Thus for a system of n particles, we have:
$\mathbf\small{\vec{L}=\vec{l}_1+\vec{l}_1+\vec{l}_1\;+\; .\; .\; .\;+\vec{l}_n=\sum\limits_{i=1}^{i=n}{\vec{l}_i}}$
• Note that, this is an addition of vectors. That is., we have to add the momenta vectorially
• In the above equation, we know that: $\mathbf\small{\vec{l}_i=\vec{r}_i \times \vec{p}_i}$
• So the equation in (1) becomes:
Eq.7.20: $\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
• If required, the $\mathbf\small{\vec{p}_i}$ can be further split up as: $\mathbf\small{\vec{p}_i=m_i \times \vec{v}_i}$
Information 3:
Eq.7.21: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_{ext}}$
• Based on this equation, we can write:
The 'rate of change of angular momentum of a system of particles' at any instant is equal to the 'sum of external torques on all the particles'
We now want to see a special case:
The angular momentum about a fixed axis
We will write it in steps:
1. From information 2 above, we have:
$\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
2. In this equation, $\mathbf\small{(\vec{r}_i\times \vec{p}_i)}$ is the angular momentum of a single particle (the ith particle)
• Also recall that the angular momentum of a single particle is denoted by '$\mathbf\small{\vec{l}}$'
• So the angular momentum of the ith particle is denoted by '$\mathbf\small{\vec{l}}$'
• That is., $\mathbf\small{\vec{l}_i=(\vec{r}_i\times \vec{p}_i)}$
3. We will consider that typical particle first.
• That is., we will find the '$\mathbf\small{\vec{l}_i}$ of a typical particle about a fixed axis' first
• Then we will sum up the angular momenta of all the particles to get the $\mathbf\small{\vec{L}}$ of the whole body
4. Consider again fig.7.79 that we saw in a previous section. It is shown again below:
• For the single particle (indicated by the red sphere), we have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
5. $\mathbf\small{\vec{r}}$ can be split up as: $\mathbf\small{\vec{r}=\vec{OC}+\vec{CP}}$
• So we get: $\mathbf\small{\vec{l}=(\vec{OC}+\vec{CP})\times \vec{p}}$
$\mathbf\small{\Rightarrow \vec{l}=\left(\vec{OC}\times \vec{p}+\vec{CP}\times \vec{p}\right)}$
• But $\mathbf\small{\vec{p}=m \times \vec{v}}$
• So we get: $\mathbf\small{\vec{l}=\left[\vec{OC}\times (m\,\vec{v})+\vec{CP}\times (m\,\vec{v})\right]}$
6. There are two terms on the right side
• Consider the first term: $\mathbf\small{\vec{OC}\times (m\,\vec{v})}$
• It is the cross product of two vectors: $\mathbf\small{\vec{OC}}$ and $\mathbf\small{(m\,\vec{v})}$
• Obviously the resulting vector will be perpendicular to $\mathbf\small{\vec{OC}}$
• The vector $\mathbf\small{\vec{OC}}$ is along the axis of rotation
• So the resulting vector from the first term is perpendicular to the axis of rotation
7. Consider the second term: $\mathbf\small{\vec{CP}\times (m\,\vec{v})}$
• $\mathbf\small{\vec{CP}}$ lies on the plane of the red circle
• $\mathbf\small{(m\,\vec{v})}$ also lies on the plane of the red circle
• So the resulting vector will be perpendicular to the plane of the red circle
• That means, the resulting vector will be parallel to the axis of rotation
8. Let us calculate this 'parallel vector' obtained from the second term: $\mathbf\small{\vec{CP}\times (m\,\vec{v})}$
• We will apply the cross product rule: $\mathbf\small{|(\vec{A}\times \vec{B})|=|\vec{A}|\times |\vec{B}|\times \sin \theta}$
• In the present case, we have:
♦ $\mathbf\small{|\vec{CP}|=r_\bot}$
♦ $\mathbf\small{|\vec{v}|=r_\bot\,|\vec{\omega}|}$
♦ Angle between the two vectors = 90o. So sin θ = sin 90 = 1
• Thus we get: $\mathbf\small{\left|\left(\vec{CP}\times (m\,\vec{v})\right)\right|=r_\bot \times (m\,r_\bot\,|\vec{\omega}|)\times 1}$
$\mathbf\small{\Rightarrow \left|\left(\vec{CP}\times (m\,\vec{v})\right)\right|=(m\,r_\bot^2\,|\vec{\omega}|)}$
9. This is the magnitude of the resulting vector of the second term
• We want the direction also
• We saw that, the resulting vector is parallel to the axis of rotation
♦ The axis of rotation is the z-axis
♦ The unit vector along the z-axis is $\mathbf\small{\hat{k}}$
• So we get:
The resulting vector from the second term is: $\mathbf\small{(m\,r_\bot^2\,|\vec{\omega}|)\hat{k}}$
10. Since this vector is parallel to the axis of rotation (the z-axis), we will denote it as $\mathbf\small{\vec{l}_z}$
• So we can write: $\mathbf\small{\vec{l}_z=(m\,r_\bot^2\,|\vec{\omega}|)\hat{k}}$
• So the result in (5) becomes: $\mathbf\small{\vec{l}=\vec{l}_z+\left(\vec{OC}\times (m\,\vec{v})\right)}$
11. This is a vector sum
• The first component $\mathbf\small{\vec{l}_z}$ is parallel to the axis of rotation
• The second component $\mathbf\small{\left(\vec{OC}\times (m\,\vec{v})\right)}$ is not parallel to the axis of rotation
• So the vector sum $\mathbf\small{\vec{l}=\vec{l}_z+\left(\vec{OC}\times (m\,\vec{v})\right)}$ is not parallel to the axis of rotation
12. We know that the angular velocity $\mathbf\small{\vec{\omega}}$ is parallel to the axis of rotation (See fig.7.78 in chapter 7.15)
• So we get an important result:
The angular velocity $\mathbf\small{\vec{\omega}}$ and angular momentum $\mathbf\small{\vec{l}}$ of a particle are not necessarily parallel
• A comparison with translational motion:
In translational motion we have:
Linear velocity $\mathbf\small{\vec{v}}$ and linear momentum $\mathbf\small{\vec{p}}$ are always parallel to each other
• Now we add the angular momenta of all the particles
• The sum will give the angular momentum $\mathbf\small{\vec{L}}$ of the whole body
14. So we can write:
$\mathbf\small{L=\sum\limits_{i=1}^{i=n}{\left(l_i \right)}}$
$\mathbf\small{\Rightarrow L=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}+\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
Where:
• $\mathbf\small{\vec{l}_{z(i)}}$ is the 'angular momentum component' of the ith parallel to the axis of rotation
• $\mathbf\small{\vec{OC}_{(i)}}$ is the vector between the two points:
♦ Origin O of the reference system
♦ Center C of the circle described by the ith particle
• $\mathbf\small{m_{(i)}}$ is the mass of the ith particle
• $\mathbf\small{\vec{v}_{(i)}}$ is the linear velocity of the ith particle
15. Thus we see that, L also has two components
Consider the first component: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$
• All vectors involved in this summation are parallel to the z-axis (axis of rotation)
• So the result of the summation will be parallel to the z-axis
• So we will denote this component as $\mathbf\small{\vec{L}_z}$
16. Consider the second component: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
• All vectors involved in this summation are perpendicular to the z-axis. We know the reason:
♦ The vector product $\mathbf\small{\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)})}$ will be perpendicular to $\mathbf\small{\vec{OC}_{(i)}}$
♦ $\mathbf\small{\vec{OC}_{(i)}}$ lies along the z-axis
• So the result of the summation will be perpendicular to the z-axis
• So we will denote this component as $\mathbf\small{\vec{L}_\bot}$
17. Thus we can write:
Eq.7.30:
$\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
• Let us consider each component separately:
We have: $\mathbf\small{\vec{L}_z=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$
• But from (10), we have: $\mathbf\small{\vec{l}_{z(i)}=(m_{(i)}\,r_{\bot (i)}^2\,|\vec{\omega}|)\hat{k}}$
• So, when we take the summation $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$,
We are actually taking the summation: $\mathbf\small{\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2\,|\vec{\omega}|)\hat{k}}}$
• $\mathbf\small{|\vec{\omega}|\hat{k}}$ is the same for all particles. So it can be taken outside
• So we get: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}=|\vec{\omega}|\,\hat{k}\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2)}}$
18. But $\mathbf\small{\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2)}}$ = I, the moment of inertia of the whole body (Eq.7.25 in chapter 7.23)
So we get: $\mathbf\small{\vec{L}_z=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}=I|\vec{\omega}|\,\hat{k}}$
19. Now consider the second component: $\mathbf\small{\vec{L}_\bot=\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
• This is a summation of vectors
• Each of those vectors results from the 'vector multiplication' of two vectors: $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
Let us see an example:
(i) Consider the square rod shown in fig.7.136 (a) below:
• It is rotating about the axis shown in blue color
• The direction of rotation is indicated by the yellow curved arrow
• It is a uniform square rod. Also, the blue axis passes through the exact center of the rod
(ii) A particle is isolated at position 'P' along the rod
• It is shown in red color
• Let us write the properties of that particle:
♦ $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
♦ $\mathbf\small{m_{(i)}=m_P}$
♦ $\mathbf\small{\vec{v}_{(i)}=\vec{v}_{(P)}}$
• So for the particle at P, we can easily calculate the cross product of $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
(iii) For the particle at P, there will be an exact replica on the other side of the axis
• This particle is at Q
(An exact replica is obtained because, the square rod is uniform and also, the axis passes through the exact center)
• The position of Q will be such that, CQ = CP
• Let us write the properties of the particle at Q:
♦ $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
♦ $\mathbf\small{m_{(i)}=m_Q}$
♦ $\mathbf\small{\vec{v}_{(i)}=\vec{v}_{(Q)}}$
• So for the particle at Q also, we can easily calculate the cross product of $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
(iv) Let us write a comparison of properties:
• For both the particles, $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
• Because of the symmetry, $\mathbf\small{m_{(P)}=m_Q}$
• Because of the symmetry, the magnitudes of $\mathbf\small{\vec{v}_{(P)}\;\;\rm{and}\;\; \vec{v}_{(Q)}}$ will be the same
♦ But their directions will be exactly opposite to each other
(v) Consider the two cross products:
(a) Cross product of $\mathbf\small{\vec{OC}_{(P)}\;\;\rm{and}\;\; (m_{(P)}\,\vec{v}_{(P)})}$
(b) Cross product of $\mathbf\small{\vec{OC}_{(Q)}\;\;\rm{and}\;\; (m_{(Q)}\,\vec{v}_{(Q)})}$
• The 'vector obtained as the cross product in (a)' will be equal and opposite to the 'vector obtained as the cross product in (b)'
• So in the summation, the two vectors will cancel each other
(vi) For each particle in the rod, there will be an exact replica on the other side of the axis
• So all particles in the rod, can be grouped into pairs such that:
• In any pair, one member is the exact replica of the other
(vii) If we draw a circle with center at C and passing through P,
♦ The point Q will also lie on the circle
♦ The point Q qill be diametrically opposite to P
• This is true for all symmetric bodies
• The members of the pairs will lie diametrically opposite on the circle with center at C
(vii) So the net result is that, the summation $\mathbf\small{\vec{L}_\bot=\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$ will become zero
• That means $\mathbf\small{\vec{L}_\bot=0}$
20. So the result in (1) becomes: $\mathbf\small{\vec{L}=\vec{L}_z}$
• But for this simplification, two conditions should be satisfied:
(i) The object must be uniform and symmetric
(ii) The rotation must be about the axis of symmetry
• For our present discussion we consider only those rotations which satisfy the two conditions
• So we can confidently use the simplified form
We can write:
Eq.7.31:
For symmetric rotation of symmetric bodies, $\mathbf\small{\vec{L}=I|\vec{\omega}|\,\hat{k}}$
The angular momentum about a fixed axis
We will write it in steps:
1. From information 2 above, we have:
$\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
2. In this equation, $\mathbf\small{(\vec{r}_i\times \vec{p}_i)}$ is the angular momentum of a single particle (the ith particle)
• Also recall that the angular momentum of a single particle is denoted by '$\mathbf\small{\vec{l}}$'
• So the angular momentum of the ith particle is denoted by '$\mathbf\small{\vec{l}}$'
• That is., $\mathbf\small{\vec{l}_i=(\vec{r}_i\times \vec{p}_i)}$
3. We will consider that typical particle first.
• That is., we will find the '$\mathbf\small{\vec{l}_i}$ of a typical particle about a fixed axis' first
• Then we will sum up the angular momenta of all the particles to get the $\mathbf\small{\vec{L}}$ of the whole body
4. Consider again fig.7.79 that we saw in a previous section. It is shown again below:
 |
| Fig.7.79 |
5. $\mathbf\small{\vec{r}}$ can be split up as: $\mathbf\small{\vec{r}=\vec{OC}+\vec{CP}}$
• So we get: $\mathbf\small{\vec{l}=(\vec{OC}+\vec{CP})\times \vec{p}}$
$\mathbf\small{\Rightarrow \vec{l}=\left(\vec{OC}\times \vec{p}+\vec{CP}\times \vec{p}\right)}$
• But $\mathbf\small{\vec{p}=m \times \vec{v}}$
• So we get: $\mathbf\small{\vec{l}=\left[\vec{OC}\times (m\,\vec{v})+\vec{CP}\times (m\,\vec{v})\right]}$
6. There are two terms on the right side
• Consider the first term: $\mathbf\small{\vec{OC}\times (m\,\vec{v})}$
• It is the cross product of two vectors: $\mathbf\small{\vec{OC}}$ and $\mathbf\small{(m\,\vec{v})}$
• Obviously the resulting vector will be perpendicular to $\mathbf\small{\vec{OC}}$
• The vector $\mathbf\small{\vec{OC}}$ is along the axis of rotation
• So the resulting vector from the first term is perpendicular to the axis of rotation
7. Consider the second term: $\mathbf\small{\vec{CP}\times (m\,\vec{v})}$
• $\mathbf\small{\vec{CP}}$ lies on the plane of the red circle
• $\mathbf\small{(m\,\vec{v})}$ also lies on the plane of the red circle
• So the resulting vector will be perpendicular to the plane of the red circle
• That means, the resulting vector will be parallel to the axis of rotation
8. Let us calculate this 'parallel vector' obtained from the second term: $\mathbf\small{\vec{CP}\times (m\,\vec{v})}$
• We will apply the cross product rule: $\mathbf\small{|(\vec{A}\times \vec{B})|=|\vec{A}|\times |\vec{B}|\times \sin \theta}$
• In the present case, we have:
♦ $\mathbf\small{|\vec{CP}|=r_\bot}$
♦ $\mathbf\small{|\vec{v}|=r_\bot\,|\vec{\omega}|}$
♦ Angle between the two vectors = 90o. So sin θ = sin 90 = 1
• Thus we get: $\mathbf\small{\left|\left(\vec{CP}\times (m\,\vec{v})\right)\right|=r_\bot \times (m\,r_\bot\,|\vec{\omega}|)\times 1}$
$\mathbf\small{\Rightarrow \left|\left(\vec{CP}\times (m\,\vec{v})\right)\right|=(m\,r_\bot^2\,|\vec{\omega}|)}$
9. This is the magnitude of the resulting vector of the second term
• We want the direction also
• We saw that, the resulting vector is parallel to the axis of rotation
♦ The axis of rotation is the z-axis
♦ The unit vector along the z-axis is $\mathbf\small{\hat{k}}$
• So we get:
The resulting vector from the second term is: $\mathbf\small{(m\,r_\bot^2\,|\vec{\omega}|)\hat{k}}$
10. Since this vector is parallel to the axis of rotation (the z-axis), we will denote it as $\mathbf\small{\vec{l}_z}$
• So we can write: $\mathbf\small{\vec{l}_z=(m\,r_\bot^2\,|\vec{\omega}|)\hat{k}}$
• So the result in (5) becomes: $\mathbf\small{\vec{l}=\vec{l}_z+\left(\vec{OC}\times (m\,\vec{v})\right)}$
11. This is a vector sum
• The first component $\mathbf\small{\vec{l}_z}$ is parallel to the axis of rotation
• The second component $\mathbf\small{\left(\vec{OC}\times (m\,\vec{v})\right)}$ is not parallel to the axis of rotation
• So the vector sum $\mathbf\small{\vec{l}=\vec{l}_z+\left(\vec{OC}\times (m\,\vec{v})\right)}$ is not parallel to the axis of rotation
12. We know that the angular velocity $\mathbf\small{\vec{\omega}}$ is parallel to the axis of rotation (See fig.7.78 in chapter 7.15)
• So we get an important result:
The angular velocity $\mathbf\small{\vec{\omega}}$ and angular momentum $\mathbf\small{\vec{l}}$ of a particle are not necessarily parallel
• A comparison with translational motion:
In translational motion we have:
Linear velocity $\mathbf\small{\vec{v}}$ and linear momentum $\mathbf\small{\vec{p}}$ are always parallel to each other
13. So we calculated the angular momentum of a single particle
• It is given by the expression in (10)
• It is given by the expression in (10)
• The sum will give the angular momentum $\mathbf\small{\vec{L}}$ of the whole body
14. So we can write:
$\mathbf\small{L=\sum\limits_{i=1}^{i=n}{\left(l_i \right)}}$
$\mathbf\small{\Rightarrow L=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}+\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
Where:
• $\mathbf\small{\vec{l}_{z(i)}}$ is the 'angular momentum component' of the ith parallel to the axis of rotation
• $\mathbf\small{\vec{OC}_{(i)}}$ is the vector between the two points:
♦ Origin O of the reference system
♦ Center C of the circle described by the ith particle
• $\mathbf\small{m_{(i)}}$ is the mass of the ith particle
• $\mathbf\small{\vec{v}_{(i)}}$ is the linear velocity of the ith particle
15. Thus we see that, L also has two components
Consider the first component: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$
• All vectors involved in this summation are parallel to the z-axis (axis of rotation)
• So the result of the summation will be parallel to the z-axis
• So we will denote this component as $\mathbf\small{\vec{L}_z}$
16. Consider the second component: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
• All vectors involved in this summation are perpendicular to the z-axis. We know the reason:
♦ The vector product $\mathbf\small{\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)})}$ will be perpendicular to $\mathbf\small{\vec{OC}_{(i)}}$
♦ $\mathbf\small{\vec{OC}_{(i)}}$ lies along the z-axis
• So the result of the summation will be perpendicular to the z-axis
• So we will denote this component as $\mathbf\small{\vec{L}_\bot}$
17. Thus we can write:
Eq.7.30:
$\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
• Let us consider each component separately:
We have: $\mathbf\small{\vec{L}_z=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$
• But from (10), we have: $\mathbf\small{\vec{l}_{z(i)}=(m_{(i)}\,r_{\bot (i)}^2\,|\vec{\omega}|)\hat{k}}$
• So, when we take the summation $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$,
We are actually taking the summation: $\mathbf\small{\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2\,|\vec{\omega}|)\hat{k}}}$
• $\mathbf\small{|\vec{\omega}|\hat{k}}$ is the same for all particles. So it can be taken outside
• So we get: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}=|\vec{\omega}|\,\hat{k}\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2)}}$
18. But $\mathbf\small{\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2)}}$ = I, the moment of inertia of the whole body (Eq.7.25 in chapter 7.23)
So we get: $\mathbf\small{\vec{L}_z=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}=I|\vec{\omega}|\,\hat{k}}$
19. Now consider the second component: $\mathbf\small{\vec{L}_\bot=\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
• This is a summation of vectors
• Each of those vectors results from the 'vector multiplication' of two vectors: $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
Let us see an example:
(i) Consider the square rod shown in fig.7.136 (a) below:
 |
| Fig.7.136 |
• The direction of rotation is indicated by the yellow curved arrow
• It is a uniform square rod. Also, the blue axis passes through the exact center of the rod
(ii) A particle is isolated at position 'P' along the rod
• It is shown in red color
• Let us write the properties of that particle:
♦ $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
♦ $\mathbf\small{m_{(i)}=m_P}$
♦ $\mathbf\small{\vec{v}_{(i)}=\vec{v}_{(P)}}$
• So for the particle at P, we can easily calculate the cross product of $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
(iii) For the particle at P, there will be an exact replica on the other side of the axis
• This particle is at Q
(An exact replica is obtained because, the square rod is uniform and also, the axis passes through the exact center)
• The position of Q will be such that, CQ = CP
• Let us write the properties of the particle at Q:
♦ $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
♦ $\mathbf\small{m_{(i)}=m_Q}$
♦ $\mathbf\small{\vec{v}_{(i)}=\vec{v}_{(Q)}}$
• So for the particle at Q also, we can easily calculate the cross product of $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
(iv) Let us write a comparison of properties:
• For both the particles, $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
• Because of the symmetry, $\mathbf\small{m_{(P)}=m_Q}$
• Because of the symmetry, the magnitudes of $\mathbf\small{\vec{v}_{(P)}\;\;\rm{and}\;\; \vec{v}_{(Q)}}$ will be the same
♦ But their directions will be exactly opposite to each other
(v) Consider the two cross products:
(a) Cross product of $\mathbf\small{\vec{OC}_{(P)}\;\;\rm{and}\;\; (m_{(P)}\,\vec{v}_{(P)})}$
(b) Cross product of $\mathbf\small{\vec{OC}_{(Q)}\;\;\rm{and}\;\; (m_{(Q)}\,\vec{v}_{(Q)})}$
• The 'vector obtained as the cross product in (a)' will be equal and opposite to the 'vector obtained as the cross product in (b)'
• So in the summation, the two vectors will cancel each other
(vi) For each particle in the rod, there will be an exact replica on the other side of the axis
• So all particles in the rod, can be grouped into pairs such that:
• In any pair, one member is the exact replica of the other
(vii) If we draw a circle with center at C and passing through P,
♦ The point Q will also lie on the circle
♦ The point Q qill be diametrically opposite to P
• This is true for all symmetric bodies
• The members of the pairs will lie diametrically opposite on the circle with center at C
(vii) So the net result is that, the summation $\mathbf\small{\vec{L}_\bot=\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$ will become zero
• That means $\mathbf\small{\vec{L}_\bot=0}$
20. So the result in (1) becomes: $\mathbf\small{\vec{L}=\vec{L}_z}$
• But for this simplification, two conditions should be satisfied:
(i) The object must be uniform and symmetric
(ii) The rotation must be about the axis of symmetry
• For our present discussion we consider only those rotations which satisfy the two conditions
• So we can confidently use the simplified form
We can write:
Eq.7.31:
For symmetric rotation of symmetric bodies, $\mathbf\small{\vec{L}=I|\vec{\omega}|\,\hat{k}}$
Let us write a summary about the above discussion:
■ In chap 7.17, we obtained the general equation for the angular momentum of any object:
Eq.7.20: $\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
■ In the present section, we applied it to objects which rotate about a fixed axis. We obtained:
Eq.7.30:
$\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
■ Also, we applied it to symmetric objects which rotate symmetrically about a fixed axis. We obtained:
Eq.7.31:
For symmetric rotation of symmetric bodies, $\mathbf\small{\vec{L}=I|\vec{\omega}|\,\hat{k}}$
■ In chap 7.17, we obtained the general equation for the angular momentum of any object:
Eq.7.20: $\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
■ In the present section, we applied it to objects which rotate about a fixed axis. We obtained:
Eq.7.30:
$\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
■ Also, we applied it to symmetric objects which rotate symmetrically about a fixed axis. We obtained:
Eq.7.31:
For symmetric rotation of symmetric bodies, $\mathbf\small{\vec{L}=I|\vec{\omega}|\,\hat{k}}$
In the next section, we will see applications of Eq.7.30
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